Problem: $r(t) = (t\sin(t), t^2)$ What is the speed of $r(t)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\sqrt{\sin^2(t) + t\sin(2t) + t^2\cos^2(t) + 4t^2}$ (Choice B) B $\sqrt{2\sin^2(t) + t\sin(2t) + t\cos(t) + 2t}$ (Choice C) C $\sqrt{\sin^2(t) + t\sin(t)\cos(t) + t^2\cos^2(t) + 4t^2}$ (Choice D) D $\sqrt{\sin^2(t) - \sin(2t) + t\cos(t) + 4t^2}$
Solution: The speed of a parametric curve is the magnitude of its velocity. If $f(t) = (a(t), b(t))$, then speed is: $\| f'(t) \| = \sqrt{ a'(t)^2 + b'(t)^2 }$ Our position function here is $r(t)$. $\begin{aligned} r'(t) &= (\sin(t) + t\cos(t), 2t) \\ \\ \text{speed} &= ||r'(t)|| \\ \\ &= \sqrt{\sin^2(t) + 2t\sin(t)\cos(t) + t^2\cos^2(t) + 4t^2} \\ \\ &= \sqrt{\sin^2(t) + t\sin(2t) +t^2\cos^2(t) + 4t^2} \end{aligned}$ Therefore, the speed of $r(t)$ is $\sqrt{\sin^2(t) + t\sin(2t) +t^2\cos^2(t) + 4t^2}$